Acid–base Titrations
Specification Reference 2.1.4 (d)–(e)
Quick Notes
- Titrations can be used to determine unknown concentrations of an acid or base
- Use mol = c × V and balanced equations
- Accurate endpoint found using an appropriate indicator
Full Notes
Titrations have been covered in more detail here later in the course and as Core Practical 2 (covered in detail here).
Titrations are used to determine the unknown concentration of an acid or alkali. The volume needed for neutralisation of a given volume of an acid or base is found using a burette (see method below).
Method:
- Use a pipette to measure a known volume of alkali (for example, NaHCO3(aq)) into a conical flask.
- Add a few drops of indicator (e.g. phenolphthalein or methyl orange).
- Titrate with acid from a burette until the colour changes at the endpoint.
- Repeat until concordant results (within 0.1 cm3) are obtained.
Calculations
Use n = c × V (with V in dm3) along with balanced equations to provide mole ratios of the acid and base being used
.Worked Example
Finding the concentration of HCl using NaOH titration
- Data
25.00 cm3 of HCl of unknown concentration is pipetted into a conical flask.
0.100 mol dm−3 NaOH is in the burette.
Concordant titre = 23.40 cm3 of NaOH to reach the endpoint. - Balanced equation
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) (1 : 1 ratio) - Moles of NaOH added
Convert volume to dm3: 23.40 cm3 ÷ 1000 = 0.02340 dm3
n(NaOH) = c × V = 0.100 × 0.02340 = 0.002340 mol - Moles of HCl in the flask
1 : 1 ratio ⇒ n(HCl) = 0.002340 mol - Concentration of HCl
Volume of HCl = 25.00 cm3 = 0.02500 dm3
c(HCl) = n ÷ V = 0.002340 ÷ 0.02500 = 0.0936 mol dm−3 - Answer
The concentration of HCl is 0.0936 mol dm−3 (3 s.f.).
Summary
- Titrations use a burette and indicator to find an unknown concentration accurately.
- Use n = c × V and balanced equations to relate volumes and concentrations.
- Obtain concordant titres within 0.1 cm3 for reliable results.